You can check that there exists a potential function, $U$, such that $\mathbf{F} = -\nabla U $, since $\nabla \wedge \mathbf{F} = 0$ (then, $\mathbf{F}$ is known as a force field). You can also verify that $U$ is given by:
$$ U(x,y) =- \frac{x^2}{y},$$
which is continuous in its domain, which is the same as the domain of $\mathbf{F}$.
Therefore, you can simply compute the work done by the (conservative) force $\mathbf{F}$ as the difference of potential between the two points $A(-1,1)$ and $B(3,2)$, that is to say:
$$\color{blue}{W_{A\to B} = U(A)-U(B)}$$
Hope this helps.
Cheers!
Some thoughts:
An example of such a force which is not defined somewhere would be the force created by a punctual mass $M$ actuating on a mass $m$, which is given by:
$$ \mathbf{F} = - G \frac{Mm}{|\mathbf{r}|^3} \mathbf{r}.$$
The gravitational potential, $\phi = G M m/|\mathbf{r}|^2$ is not defined at $(x,y,z) = (0,0,0)$ but we can still compute the work done by gravity when actuating along a certain path. We call this potential energy.